/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 A couple is traveling by car dow... [FREE SOLUTION] | 91Ó°ÊÓ

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A couple is traveling by car down a straight highway at \(40 \mathrm{~km} / \mathrm{h}\). They see an accident in the distance, so the driver applies the brakes, and in \(5.0 \mathrm{~s}\) the car uniformly slows down to rest. (a) The direction of the acceleration vector is (1) in the same direction as, (2) opposite to, (3) at \(90^{\circ}\) relative to the velocity vector. Why? (b) By how much must the velocity change each second from the start of braking to the car's complete stop?

Short Answer

Expert verified
(a) Acceleration is opposite to the velocity vector. (b) Velocity changes by -2.222 m/s each second.

Step by step solution

01

Analyze Initial Conditions

The initial velocity of the car is \(40 \text{ km/h}\). This needs to be converted into meters per second for standard units. To do this, use the conversion factor: \(1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\). Therefore, the initial velocity \(v_i = 40 \times \frac{1}{3.6} \text{ m/s} = 11.11 \text{ m/s}\).
02

Determine the Final Velocity

When the car comes to rest, the final velocity \(v_f = 0 \text{ m/s}\).
03

Calculate Time Interval

The time duration from the moment the driver applies the brakes to the car coming to a complete stop is given as \(5.0 \text{ s}\).
04

Identify Direction of Acceleration

Acceleration is the rate of change of velocity. Since the car is slowing down, the acceleration is in the opposite direction to the initial velocity, which means the answer to part (a) is (2) opposite to the velocity vector.
05

Calculate Change in Velocity per Second

Velocity change is uniform over time. The total change in velocity is \(\Delta v = v_f - v_i = 0 \text{ m/s} - 11.11 \text{ m/s} = -11.11 \text{ m/s} \). Since this change happens uniformly over \(5.0 \text{ s}\), the change in velocity per second, i.e., acceleration \(a\), is \( \frac{-11.11 \text{ m/s}}{5 \text{ s}} = -2.222 \text{ m/s}^2 \). The car's velocity changes by \(-2.222 \text{ m/s}\) each second.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematics
Kinematics is a fundamental branch of physics that deals with the motion of objects. It focuses on various aspects such as position, velocity, and acceleration without considering the causes of the motion, which are often explained by dynamics. In this exercise, the car's motion is an excellent example of kinematics. The car is moving in a straight line, and the concepts of initial velocity, final velocity, and time duration are explored.

Understanding kinematics requires familiarity with a few important parameters:
  • Displacement - The change in position of an object.
  • Velocity - Describes speed and direction of the moving object.
  • Acceleration - The rate at which velocity changes with time.
The exercise specifically revolves around uniform acceleration, where the car decelerates at a constant rate. This uniform change allows us to use basic kinematic equations to determine important factors such as velocity change over time.
Velocity Conversion
Velocity conversion is crucial when dealing with different units of measurement in physics. In the original exercise, the car's initial velocity is given in kilometers per hour (km/h), but standard physics calculations require units in meters per second (m/s).
To convert velocity from km/h to m/s, a conversion factor of \(1 \, \text{km/h} = \frac{1}{3.6} \, \text{m/s}\) is used. This reflects the conversion between kilometers to meters and hours to seconds. Applying this conversion gives an initial velocity \(v_i\) of 11.11 m/s from an original 40 km/h. Understanding how to switch between these units ensures precision in calculations, especially when applying formulas that require consistent units, such as those in kinematics equations.
Acceleration Direction
The direction of acceleration is an important concept in understanding how forces affect motion. In kinematics, acceleration indicates how an object's velocity changes over time. For this car scenario, once the brakes are applied, the velocity changes directionally as the car slows down. Since the car is slowing to a stop, the acceleration is opposite to the direction of its motion. This is because acceleration acts to reduce the velocity over time. Thus, in the exercise, the acceleration vector points opposite to the initial direction of motion.

This is an essential concept in analyzing situations involving moving objects where forces cause them to decelerate. Grasping this principle allows us to predict not only the magnitude of change but the orientation of the movement vector as well.
Velocity Change
The change in velocity during uniform acceleration can be calculated using kinematic equations. In this exercise, the car's velocity changes uniformly as it slows down over a span of time due to braking. The total velocity change \(\Delta v\) is the initial velocity minus the final velocity, represented by:\[ \Delta v = v_f - v_i \]where \(v_f\) is the final velocity (0 m/s when stopped) and \(v_i\) is the initial velocity (11.11 m/s). Substituting these values gives a velocity change of -11.11 m/s. To determine how much the velocity changes each second, divide the total change by the time duration:\[ a = \frac{-11.11 \, \text{m/s}}{5 \, \text{s}} \]This gives an acceleration of -2.222 m/s², interpreted as a decrease of 2.222 m/s in velocity each second. Understanding this daily-life application of physics enhances comprehension of how changes in motion manifest in uniform circumstances.

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Most popular questions from this chapter

A rifle bullet with a muzzle speed of \(330 \mathrm{~m} / \mathrm{s}\) is fired directly into a special dense material that stops the bullet in \(25.0 \mathrm{~cm}\). Assuming the bullet's deceleration to be constant, what is its magnitude?

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